Asked by loulou
If 15.0 grams of C3H6, 10.0 grams of oxygen, and 5.00 grams of NH3 are reacted, what mass of acrylonitrile
can be produced assuming 100% yield, what mass of the excess reactants remain?
15g C3H6 (1 mol / 42.08 g) = .35 mol C3H6
.36 mol C3H6 ( 2 mol C3H3N / 2 mol C3H6) = .36 mol C3H3N
10g O2 ( 1 mol/ 32 g O2) = .313 mol O2
.313 mol O2 (2 mol C3H3N/ 3 mol O2) -= .209 mol C3H3N which is the limiting reactant
5 g NH3 ( 1 mol NH3/ 17.03 g) = .294 mol C3H3N
.209 mol C3H3N ( 32 c3H6 / 2 mol C3H3N) = .209 m C3H6
.36mol - .209 mol = .151 mol C3H6
.151 mol C3H6 (42.08 g / 1 mol C3H6 ) = 6.35 g C3H6
Thank you for your help!!
can be produced assuming 100% yield, what mass of the excess reactants remain?
15g C3H6 (1 mol / 42.08 g) = .35 mol C3H6
.36 mol C3H6 ( 2 mol C3H3N / 2 mol C3H6) = .36 mol C3H3N
10g O2 ( 1 mol/ 32 g O2) = .313 mol O2
.313 mol O2 (2 mol C3H3N/ 3 mol O2) -= .209 mol C3H3N which is the limiting reactant
5 g NH3 ( 1 mol NH3/ 17.03 g) = .294 mol C3H3N
.209 mol C3H3N ( 32 c3H6 / 2 mol C3H3N) = .209 m C3H6
.36mol - .209 mol = .151 mol C3H6
.151 mol C3H6 (42.08 g / 1 mol C3H6 ) = 6.35 g C3H6
Thank you for your help!!
Answers
Answered by
DrBob222
The first thing you need to do is write and balance the equation. then I can help.
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